This is a binomial theorm problem
asked Jun 26, 2013 in Algebra 1 Answers by anonymous

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(2^n)-1=1+2+2^2+2^3+...+2^(n-1) so 1-2^-n=(1+2+2^2+2^3+...+2^(n-1))/2^n

Geometrical progression S(n-1)=1+2+2^2+...+2^(n-1); r=2 and a=1 (∑r^k for 0≤k≤n-1).

2S(n-1)=2+2^2+2^3+...+2^n; 2S(n-1)-S(n-1)=S(n-1)=2^n-1.

Also, x^n-1=(x-1)(1+x+x^2+...+x^(n-1)). When x=2, 2^n-1=(1+2+2^2+2^3+...+2^(n-1)).

2=1/(1-x) where x=1/2.

So, using binomial expansion, 2=(1-x)^-1=1+x+x^2+x^3+... =1+1/2+1/4+1/8+... =

(2^n+2^(n-1)+2^(n-2)+...+2^3+2^2+1)/2^n where n tends to infinity.


answered Jun 21, 2016 by Rod Top Rated User (429,280 points)
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